Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Ed Sandifer writes on Euler's wonderful work on prime numbers: The sum of the series of reciprocals of prime numbers $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...$ is infinitely large, and is infinitely less than the harmonic series $1 +\frac{1}{2}+\frac{1}{3}+...$. Moreover, the first sum is almost the logarithm of the second sum. Euler’s proof of this again on adding and subtracting series that do not converge, and so does not meet modern standards of rigor. Unfortunately, Euler’s proof can’t really be “repaired”. http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2029%20infinitely%20many%20primes.pdf

Is it really true that this proof can't be repaired with modern tools? If it can be repaired, however, what has to be done?

share|improve this question
Why 4 votes to close (at the time I write this)? –  KConrad Jun 29 '13 at 22:56
I don't know. No one has explained in the comments. –  S. Carnahan Jun 29 '13 at 22:59
add comment

2 Answers 2

up vote 17 down vote accepted

This result of Euler is the last theorem in http://eulerarchive.maa.org/docs/originals/E072.pdf, and if you prefer English to Latin look at the last theorem in http://eulerarchive.maa.org/docs/translations/E072en.pdf. I don't know why Sandifer says the proof can't be repaired. Euler's goal is to show $\sum_{p} 1/p$ diverges, and that in some sense $\sum_{p \leq n} 1/p$ diverges like $\log(\log n)$. (Euler didn't speak directly about asymptotic approximations, and he wrote the last conclusion as $\sum_p 1/p = l(l(\infty))$, using $l$ for the natural logarithm.)

Euler's proof: for each positive integer $k$, set $A_k = \sum_p 1/p^k$. Then $$ e^{\sum_{k \geq 1} A_k/k} = \sum_{n \geq 1} \frac{1}{n} $$ and $\sum_{k \geq 2} A_k/k < \infty$, so $A_1$, which is $\sum_p 1/p$, diverges. Since $A_1 = \infty$ while $\sum_k A_k/k < \infty$, we have $$ e^{A_1} = \sum_{n \geq 1} \frac{1}{n}, $$ so $$ \sum_p \frac{1}{p} = A_1 = \log\left(\sum_{n \geq 1} \frac{1}{n}\right). $$ QED

The last part is kind of a bogus calculation, but the rest can be fixed using the zeta-function.

  1. In place of $A_k$ use $\sum_{p} 1/p^{ks}$, so the replacement of the first displayed equation above is $$ e^{\sum_{p,k} 1/kp^{ks}} = \sum_{n \geq 1} \frac{1}{n^s} $$ for ${\rm Re}(s) > 1$. In other words, this is saying $\sum_{p,k} 1/kp^{ks}$ is a logarithm of $\zeta(s)$ for ${\rm Re}(s) > 1$. Series are absolutely convergent, so all manipulations of terms in the series are justified. Euler does all his work at $s = 1$, where expressions may be divergent.

  2. Euler's claim that $\sum_{k \geq 2} A_k/k < \infty$ is correct, and the modern replacement for this is the more general observation that $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$. Euler's statement is at $s=1$.

  3. Let $s$ be real and greater than 1, so the displayed equation can be rewritten as $$ \sum_{k \geq 1} \sum_{p} \frac{1}{kp^{ks}} = \log\left(\sum_{n \geq 1} \frac{1}{n^s}\right). $$ As $s \rightarrow 1^+$, $\sum_n 1/n^s \rightarrow \infty$, so the right side above tends to $\infty$ as $s \rightarrow 1^+$. On the left side above, split off the terms where $k=1$ from the terms where $k \geq 2$. Since $\sum_{k \geq 2} \sum_p 1/kp^{ks}$ is convergent for ${\rm Re}(s) > 1/2$, this series has a limit as $s \rightarrow 1^+$ (Dirichlet series are continuous on open half-planes where they converge). Therefore the sum of the terms with $k = 1$, $\sum_p 1/p^s$, must tend to $\infty$ as $s \rightarrow 1^+$. For $s > 1$ we have $\sum_p 1/p > \sum_p 1/p^s$, and the right side tends to $\infty$ as $s \rightarrow 1^+$, so $\sum_p 1/p = \infty$.

share|improve this answer
As far as concluding $\sum 1/p = \infty$ goes, I don't understand what's wrong with Euler's original argument. If $A_1 = \sum 1/p$ was really finite, then nothing would stop us from expanding the finite quantity $\mathrm{exp}(\sum A_k / k)$ and getting $\sum 1/n$, contradiction. –  Vivek Shende Jun 30 '13 at 2:36
@VivekShende: That's true. Since everything in sight is positive, convergence of $A_1$ would imply convergence of the harmonic series. –  KConrad Jun 30 '13 at 2:41
Well, strictly speaking, by rearranging terms you'd get $\exp(\sum A_k/k) = \prod_p \exp(\sum 1/kp^k) = \prod 1/(1-1/p)$, which is $\sum 1/n$ by one argument or another. –  KConrad Jun 30 '13 at 2:47
add comment

The proof is easily repaired (although I'm not sure exactly what the details of the proof you're referring to are and don't know how closely you require the repairing of a proof to adhere to the original proof). Observe that

$$\prod_{p \le n} \left( \frac{1}{1 - \frac{1}{p}} \right) \ge \sum_{k=1}^n \frac{1}{k}.$$

Now take logarithms, use the fact that $\log \frac{1}{1 - x} = x + O(|x|^2)$, and let $n \to \infty$.

share|improve this answer
It is not clear to me that the OP even knows what they mean by repairing the proof, since the source the OP is using does not actually give the proof itself. This is as good a meaning as any. –  Qiaochu Yuan Jun 29 '13 at 21:40
I think the question asked is pretty clear, and can be addressed by looking at the proof that Euler wrote (not just at what Sandifer wrote). –  KConrad Jun 30 '13 at 0:38
Okay, I'm removing my previous comments. I suppose this is a situation where I was supposed to use the new "messaging" system. –  S. Carnahan Jun 30 '13 at 3:31
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.